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In the figure above, points P and Q lie on the circle with

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In the figure above, points P and Q lie on the circle with [#permalink] New post 12 Sep 2005, 08:02
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A
B
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D
E

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In the figure above, points P and Q lie on the circle with center O. What is the value of s?

A. \(\frac{1}{2}\)

B. \(1\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. \(\frac{\sqrt{2}}{2}\)
[Reveal] Spoiler: OA

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Re: gmatprep geo [#permalink] New post 23 Aug 2009, 14:54
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Check another approach for this problem.
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Re: gmatprep geo [#permalink] New post 06 Sep 2009, 12:14
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There's a simple reason and misconception as to why people find this question confusing:
--> the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles.
--> that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin!

However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer!

Upon closer inspection of the question:

P(-sq(3),1)
Which means the triangle beneath point P has:
height = 1
base = sq(3)
meaning >> radius = 2 (30-60-90 triangle)

This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1!

**Have a look at the attachment and then it should make perfect sense!
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Re: coordinate geometry [#permalink] New post 16 Jun 2009, 09:18
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The answer is B.

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \(\triangle NOP\). We are given the co-ordinates of P as \((-\sqrt{3}, 1)\). i.e. \(ON=\sqrt{3}\) and \(PN=1\). Also we know angle \(PNO=90\textdegree\). If you notice this is a \(30\textdegree-60\textdegree-90\textdegree\)triangle with sides \(1-sqrt3-2\) and angle \(NOP=30\textdegree\).

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a right angle triangle \(\triangle QOR\). We know angle \(QRO=90\textdegree\).

Now
angle NOP + angle POQ (= 90 -- given) + angle QOR = 180

\(\Rightarrow 30\textdegree+90\textdegree+angleQOR=180\)

\(\Rightarrow angleQOR=60\)

If you notice \(\triangle QOR\) is also a \(30\textdegree-60\textdegree-90\textdegree\)triangle with sides \(1-sqrt3-2\) and angle \(OQR=30\textdegree\)

The side opposite to this angle is OR = 1.

Hence answer is B.
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 [#permalink] New post 17 Sep 2005, 21:41
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there is no need for much calculation,
simply switch their X and Y

P and Q are always perpendicular and same distance from O

OP is perpendicular to OQ means you rotate OQ 90 degrees, you will
get OP, in terms of X,Y value, that is just to switch them and flip the sign
to have (-Y,X)
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Re: GMAT Prep 1 [#permalink] New post 27 Apr 2008, 21:49
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jbpayne wrote:
Thats what I thought, but the answer is B, 1. Anyone able to figure this one out?


There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 05 Jan 2011, 10:46
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The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:

Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?
Attachment:
Ques1.jpg
Ques1.jpg [ 7.87 KiB | Viewed 8985 times ]

Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line \(= \sqrt{3}\) cm and blue line = 1 cm

So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
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Re: GMAT Prep 1 [#permalink] New post 27 Apr 2008, 22:08
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bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1


I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 27 Apr 2010, 15:02
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zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\);
Point P in II quadrant \((-\sqrt{3}, 1)\);
Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP);
Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 04 Jan 2011, 17:37
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tonebreeze: As requested, I am giving you my take on this problem (though I guess you have already got some alternative solutions that worked for you)

I would suggest you first go through the explanation at the link given below for another question. If you understand that, solving this question is just a matter of seconds by looking at the diagram without using any formulas.
http://gmatclub.com/forum/coordinate-plane-90772.html#p807400

P is (\(-\sqrt{3}\), 1). O is the center of the circle at (0,0). Also OP = OQ. When OP is turned 90 degrees to give OQ, the length of the x and y co-ordinates will get interchanged (both x and y co-ordinates will be positive in the first quadrant). Hence s, the x co-ordinate of Q will be 1 and y co-ordinate of Q will be \(\sqrt{3}\).
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 [#permalink] New post 18 Sep 2005, 11:20
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Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....

Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 11 Oct 2009, 11:48
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Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. -> (1)
Also from the right triangle, we know the length of PQ=2\sqrt{2}.
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

From (1) and (2) you can solve for s.
Hope this helps.
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Re: Help on a Geometry PS Question [#permalink] New post 09 Sep 2010, 15:46
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Attachment:
a.jpg
a.jpg [ 12.91 KiB | Viewed 46384 times ]


Line OP => y=mx => 1 = \(-sqrt{3} * m\)

=> slope of OP = \(\frac{-1}{sqrt{3}}\)

Since OP is perpendicular to OQ => m1*m2 = -1
=> slope of OQ = \(sqrt{3}\)

equation of line OQ => y = \(sqrt{3}x\)

since s and t lie on this line t= \(sqrt{3}s\)

also radius of circle = \(sqrt{ 1^2+ 3}\) = 2 = \(sqrt{ s^2+ t^2}\)

=> \(4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2\)
=> s=1
----------------------------------------------------------------------

Attachment:
b.jpg
b.jpg [ 13.36 KiB | Viewed 46336 times ]


Using Similar triangles.
In triangle APO , angle APO + AOP = 90 ---------------------1
In triangle BOQ , angle BOQ + BQO= 90 ---------------------2

Also SINCE angle POQ = 90 as given.

angle AOP + angle BOQ = 90 ------------------3

Using equations 1 and 3 we get Angle APO = Angle BQO -----------4

using equation 2 ,3, 4
we get Angle APO = Angle BQO and AOP = OQB

Since hypotenuse is common => both the triangle are congruent

=> base of triangle APO = height of triangle AQO and vise versa.

Thus s = y co-ordinate of P = 1
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Last edited by gurpreetsingh on 09 Sep 2010, 16:00, edited 1 time in total.
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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 23:55
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agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \(\sqrt{3}\)
PA =1
Thus radius, OP = 2

Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2 (radius)
Therefore BQ = \(\sqrt{3}\)
and OB = 1

We need value of r which is BQ = \(\sqrt{3}\)

Ans D it is!
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Re: GMAT Prep 1 [#permalink] New post 29 Apr 2008, 08:36
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sondenso wrote:
bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1


I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1




From the picture t^2 +s^2 = 4(radius = 2)
By substituting t =√3s we get s=1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 09 Nov 2009, 19:25
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Correct me if I'm wrong, (and forgive my english)
the coordinates of a point p(x,y) that lays on a circle can be expressed as p(sen,cos).
we know that sen π/6 = cos π/3,
and that sen π/3 = cos π/6

in this case I just have to invert x with y and y with x -> indeed we have p(y,x)

As we know x and y have to be positive we have to change the sign for x, the new coordinates are p (y,-x).

in this way to solve the problem doesn't take more than 20 secs.
hope I was enough clear.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 27 Apr 2010, 14:36
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I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
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Re: gmatprep geo [#permalink] New post 20 Oct 2013, 19:35
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obs23 wrote:
chetanojha wrote:
Check another approach for this problem.


Can we assume that coordinates of the center O are (0,0) here, can we always assume so?


From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 13 Oct 2009, 04:29
DenisSh wrote:
Sorry, didn't get that part:
badgerboy wrote:
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)


The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2

from Right angle Triangle POQ we had got the length of PQ^2 as 8
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 13 Oct 2009, 12:15
Yes Answer is 1.

S^2 + T^2 = 4 --> equation 1

(t-1)^2 + (s-(-sqrt(3))^2 = 2^2 + 2^2 = 8 --> equation 2

solving for s gives 1
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Re: Plane Geometry, Semicircle from GMATPrep   [#permalink] 13 Oct 2009, 12:15

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